few_shots: Example:
Problem/activity:
Task
Cruz and Erica were both getting ready for soccer.
Cruz ran 1 lap around the school.
Erica ran 3 laps around the playground.
Erica said,
I ran more laps, so I ran farther.
Cruz said,
4 laps around the school is 1 mile, but it takes 12 laps around the playground to go 1 mile. My laps are much longer, so I ran farther.
Who is right? Draw a picture to help you explain your answer.


Standard options: 
A. Explain why a fraction a/b is equivalent to a fraction (n × a)/(n × b) by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions.
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.
B. Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.


Thought: This task asks students to compare two fractions that are equal. The two fractions have different numerators and denominators (e.g. 1/4 and 3/12). The students need to explain why the two fractions are equal by drawing a picture, which involves using visual fraction models.
Answer: A, B

Now, assign the following problem or activity to one or more relevant standards it aligns with.
shot_info: im_task_000212

4
options: Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model.

Use permutations and combinations to compute probabilities of compound events and solve problems.

Apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model.

Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A, and interpret the answer in terms of the model.
problem_activity: Optional activity
The mathematical purpose of this activity is for students to apply conditional probability in the context of a problem. It is important that students use the ruler to draw the X on each card so that slight differences in the way the Xs are drawn are not recognizable.
Launch
Arrange students in groups of 2. Give 3 cards and one ruler to each group.
Demonstrate the game by playing one round with the students.
Tell students, "The first card will be blank on both sides, the second card will have an X on one side, and the third card will have an X on both sides. It is important to use the ruler when drawing the Xs on the cards since you should not be able to guess the card based on small differences in how the X is drawn."
Place all three cards in the bag. Remove one and hold it against the wall so that only one side can be seen by the students. Ask them which card they think this card is based on what they are seeing. Is it the blank card, the card with one X on one side, or the card with an X on both sides?
Engagement: Provide Access by Recruiting Interest.
Begin with a small-group or whole-class demonstration of how to play the game. Check for understanding by inviting students to rephrase directions in their own words.
Supports accessibility for: Memory; Conceptual processing
Student Facing
Your teacher will give you 3 index cards.
Leave one card blank.
Use a ruler to draw lines connecting opposite corners to make an X on one side of the second card.
Use a ruler to draw lines connecting opposite corners to make an X on both sides of the third card.
Put all three cards in the bag.
One partner will remove a card from the bag and place it on the desk immediately so that only one side of the card can be seen. The goal is to guess correctly which card is on the desk: the blank card, the card with an X only on one side, or the card with an X on both sides.
Noah is playing the game and is looking at a card that shows an X. He says, “I have a fifty-fifty chance of correctly guessing which card it is.” Do you agree with Noah? Explain your reasoning.
Play the game many times with your partner, taking turns for who takes a card out of the bag. Record whether the side that shows has an X or is blank, which card you guess, and which card it actually is when you check. Continue to play until your teacher tells you to move on.
Use the results from your games to estimate
\(P(\text{the card had an X on both sides | the side showing had an X on it})\)
. Explain or show your reasoning.
Since all of the outcomes are known, find the actual probability
\(P(\text{the card had an X on both sides | the side showing had an X on it})\)
. Explain or show your reasoning.
Student Response
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Anticipated Misconceptions
Some students may have difficulty figuring out the answer to question 4. Prompt students to think about the total number of sides rather than the number of cards. Emphasize that the number of sides are the sample space, not the number of cards.
Activity Synthesis
The purpose of this discussion is for students to communicate their reasoning with each other.
Poll the class to collect information on about what was shown and which card it turned out to be.
|                  | both sides blank | one X | two Xs |
|------------------|------------------|-------|--------|
| blank side shown |                  |       |        |
|     X shown      |                  |       |        |
Students should notice that two of the positions (top right and bottom left) must contain zeros.
Here are some questions for discussion.
“How did you figure out the answer to question 4?” (I figured out that it was
\(\frac{2}{3}\)
since
\(P(\text{X on both sides | saw an X}) = \frac{P(\text{X on both sides and saw an X})}{P(\text{saw an X})}\)
. We know
\(P(\text{the card had an X on both sides and the side showing had an X on it}) = \frac{2}{6}\)
(there are 6 sides that could be showing and 2 of them are Xs from the card with an X on both sides) and
\(P(\text{the side showing had an X on it}) = \frac{3}{6}\)
(there are 6 sides that could be showing and 3 of them have an X). Then we can find
\(P(\text{the card had an X on both sides | the side showing had an X on it}) = \frac{ \left( \frac{2}{6} \right)}{\left( \frac{3}{6} \right)}\)
which is
\(\frac{2}{3}\)
.
“Do you think that playing the game 100 times in question 2 would have helped you arrive at the answer to question 4? Explain your reasoning.” (I think it might have made me feel more confident in my answer to question 4. Since my group only played the game 12 times, it really did not give me much insight. I wonder what would have happened if we played it 100 times. Maybe we could combine the data from the whole class to get a better ideas of what would happen in the long run.)
Speaking: MLR8 Discussion Supports.
Use this routine to support whole-class discussion. For each observation that is shared, ask students to restate what they heard using precise mathematical language. Consider providing students time to restate what they hear to a partner before selecting one or two students to share with the class. Ask the original speaker if their peer was accurately able to restate their thinking. Call students' attention to any words or phrases that helped to clarify the original statement. This provides more students with an opportunity to produce language as they interpret the reasoning of others.
Design Principle(s): Support sense-making


correct_option_index: 3